# How do you find all the asymptotes for function f(x)= (x+3)/(x^2-9)?

Jul 19, 2015

$f \left(x\right) = \frac{x + 3}{{x}^{2} - 9} = \frac{x + 3}{\left(x - 3\right) \left(x + 3\right)} = \frac{1}{x - 3}$

with exclusion $x \ne - 3$.

Hence the asymptotes are $y = 0$ and $x = 3$.

#### Explanation:

Use the difference of squares identity: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = x$ and $b = 3$ to find:

$\left({x}^{2} - 9\right) = \left({x}^{2} - {3}^{2}\right) = \left(x - 3\right) \left(x + 3\right)$

So:

$f \left(x\right) = \frac{x + 3}{{x}^{2} - 9} = \frac{x + 3}{\left(x - 3\right) \left(x + 3\right)} = \frac{1}{x - 3}$

with exclusion $x \ne - 3$.

As $x \to \pm \infty$ we find $f \left(x\right) \to 0$, so $y = 0$ is a horizontal asymptote.

There is a vertical asymptote at $x = 3$ since the denominator becomes zero and the numerator is non-zero.